Element has Idempotent Power in Finite Semigroup

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Let $\left({S, \circ}\right)$ be a finite semigroup.

For every element in $\left({S, \circ}\right)$, there is a power of that element which is idempotent.

That is:

$\forall x \in S: \exists i \in \N: x^i = x^i \circ x^i$


From Finite Semigroup Equal Elements for Different Powers, we have:

$\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$

Let $m > n$.

Let $n = k, m = k + l$.


$\forall x \in S: \exists k, l \in \N: x^k = x^{k + l}$

Now we show that:

$x^k = x^{k + l} \implies x^{k l} = x^{k l} \circ x^{k l}$

That is, that $x^{k l}$ is idempotent.


\(\displaystyle x^k\) \(=\) \(\displaystyle x^{k + l}\)                    
\(\displaystyle \implies\) \(\displaystyle x^k \circ x^l\) \(=\) \(\displaystyle x^{k + l} \circ x^l\)          Both sides ${} \circ x^l$          
\(\displaystyle \implies\) \(\displaystyle x^{k + l}\) \(=\) \(\displaystyle x^{k + 2 l}\)          Index Laws for Semigroup‎          
\(\displaystyle \implies\) \(\displaystyle x^k\) \(=\) \(\displaystyle x^{k + 2 l}\)          as $x^k = x^{k + l}$          

From here we can easily prove by induction that:

$\forall n \in \N: x^k = x^{k + n l}$

In particular:

$x^k = x^{k + k l} = x^{k \left({l + 1}\right)}$

There are two cases to consider:

$(1): \quad$ If $l = 1$, then $x^k = x^{k \left({l + 1}\right)} = x^{2 k} = x^k \circ x^k$, and $x^{k l} = x^k$ is idempotent.
$(2): \quad$ If $l > 1$, then:
\(\displaystyle x^k\) \(=\) \(\displaystyle x^{k + kl}\)                    
\(\displaystyle \) \(=\) \(\displaystyle x^{k \left({l + 1}\right)}\)                    
\(\displaystyle \implies\) \(\displaystyle x^k \circ x^{k \left({l - 1}\right)}\) \(=\) \(\displaystyle x^{k \left({l + 1}\right)} \circ x^{k \left({l - 1}\right)}\)          Both sides ${} \circ x^{k \left({l - 1}\right)}$          
\(\displaystyle \implies\) \(\displaystyle x^{k + k l - k}\) \(=\) \(\displaystyle x^{kl + k + k l - k}\)          Index Laws for Semigroup‎          
\(\displaystyle \implies\) \(\displaystyle x^{k l}\) \(=\) \(\displaystyle x^{k l + k l}\)                    
\(\displaystyle \) \(=\) \(\displaystyle x^{k l} \circ x^{k l}\)                    

... and again, $x^{k l}$ is idempotent.